Transcript Entropy, Free Energy, and Equilibrium

Entropy, Free
Energy, and
Equilibrium
Entropy 2009-2010
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Spontaneous Processes and
Entropy
One of the main objectives in studying
thermodynamics, as far as chemists
are concerned, is to be able to predict
whether or not a reaction will occur
when reactants are brought together
under a special set of conditions (for
example, at a certain temperature,
pressure, and concentration).
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A reaction that does occur under the
given set of conditions is called a
spontaneous reaction. If a reaction
does not occur under specified
conditions, it is said to be
nonspontaneous.
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We observe spontaneous physical and chemical
processes every day, including many of the
following examples:
A waterfall runs downhill, but never up,
spontaneously.
A lump of sugar spontaneously dissolves in a
cup of coffee, but dissolved sugar does not
spontaneously reappear in its original form.
Water spontaneously freezes below 0 oC, and
ice melts spontaneously above 0 oC (at 1 atm).
Iron exposed to water and oxygen
spontaneously forms rust, but rust does not
spontaneously change back to iron.
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These examples show that processes
that occur spontaneously in one
direction cannot, under the same
conditions, also take place
spontaneously in the opposite
direction.
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If we assume that spontaneous
processes occur so as to decrease the
energy of a system, we can explain
why a ball rolls downhill and why
springs in a clock unwind. Similarly, a
large number of exothermic reactions
are spontaneous. An example is the
combustion of methane
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
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DH = -890.4 kJ
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But consider a solid-to-liquid phase
transition such as this spontaneous
process that occurs above 0 oC
H2O(s)
H2O(l)
DH = 6.01 kJ
In this case, the assumption that
spontaneous process always decrease
a system’s energy fails.
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Another example that contradicts our
assumption is the dissolution of
ammonium nitrate in water:
NH4NO3(s)
H2 O
NH4+(aq) + NO3-(aq)
DH = 25 kJ
This process is spontaneous, and yet
it is also endothermic.
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From the study of the examples mentioned and
many more cases, we come to the following
conclusion: Exothermicity favors the
spontaneity of a reaction but does not
guarantee it. Just as it is possible for an
endothermic reaction to be spontaneous, it is
possible for an exothermic reaction to be
nonspontaneous. In other words, we cannot
decide whether or not a chemical reaction
will occur spontaneously solely on the basis
of energy changes in the system. To make
this kind of prediction we need another
thermodynamic quantity, which turns out to
be entropy.
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Entropy
In order to predict the spontaneity of
a process, we need to know two things
about the system. One is the change
in enthalpy, DH. The other is change
in entropy, (DS). Entropy is a measure
of the randomness or disorder of a
system. The greater the disorder of
a system, the greater its entropy.
Conversely, the more ordered a
system, the smaller its entropy.
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For any substance, the particles in the
solid state are more ordered than
those in the liquid state, which in turn
are more ordered than those in the
gaseous state. So for the same molar
amount of a substance, we can write
Ssolid < Sliquid << Sgas
In other words, entropy describes the
extent to which atoms, molecules, or
ions are distributed in a disorderly
fashion in a given region of space.
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It is possible to determine the absolute
entropy of a substance, something we
cannot do for enthalpy. According to the
third law of thermodynamics, the entropy
of a perfect crystalline substance is zero
at absolute zero (0 K). If the crystal is
impure or if it has defects, then its
entropy is greater than zero even at 0 K
because it would not be perfectly
ordered.
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The important point about the third law
is that it allows us to determine the
absolute entropies of substances.
The standard entropies (So) that are
listed on your reference sheets are
the absolute entropies of substances
at 1 atm and 25 oC. These are the
values that are generally used in
calculations.
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The units of entropy are J/K or
J/K.mol for 1 mole of the
substance. We use joules rather
than kilojoules because entropy values
are typically quite small.
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Consider a certain process in which a
system changes from some initial
state to some final state. The
entropy change for the process, DS,
is
DS = Sfinal - Sinitial
If the change results in an increase
in randomness or disorder, then the
change in entropy is positive (+DS)
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At absolute zero, a substance
has a zero entropy value
(assuming that it is a perfect
crystalline solid). As it is
heated, its entropy increases
gradually because of greater
molecular motion. At the
melting point, there is a
sizable increase in entropy as
the more random liquid state
is formed. Further heating
increases the entropy of the
liquid again due to enhanced
molecular motion. At the
boiling point there is a large
increase in entropy as a
result of the liquid to gas
transition. Beyond that
temperature, the entropy of
the gas continues to rise with
increasing temperature.
Entropy increase of a substance as the
temperature rises from absolute zero.
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Processes that
lead to an
increase in the
entropy of the
system include
•melting
•vaporization
•dissolving*
•heating
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*Dissolving ionic compounds in water does not
always result in an increase in entropy. For
ionic compounds that contain Al+3 or Fe+3,
hydration (the process of surrounding the
ions by water molecules) can increase the
order of the water molecules so much that
the entropy change for the overall process
can actually be negative (-DS).
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The Second Law of Thermodynamics
The connection
between entropy and
the spontaneity of a
reaction is
expressed by the
second law of
thermodynamics:
the entropy of the
universe increases
in a spontaneous
process
DSuniverse = DSsystem + DSsurroundings > 0
and remains
DSuniverse = DSsystem + DSsurroundings = 0
unchanged in an
equilibrium process.
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For a spontaneous process, the second law says
that DSuniv must be greater than zero, but it
does not place a restriction on either DSsys or
DSsurr. Thus it is possible for either DSsys or
DSsurr to be negative, as long as the sum of
these two quantities is greater than zero.
• What if for some process we find that DSuniv
is negative?
The reaction is spontaneous in the opposite
direction.
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To calculate DSuniv, we need to know
both DSsys and DSsurr.
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Entropy Changes in the System
Suppose that the system is represented by the
following reaction:
aA + bB cD + dD
As in the case for enthalpy of a reaction, the
standard entropy of reaction, DSorxn is given by
DSorxn = [cSo(C) + dSo(D)] - [aSo(A) + bSo(B)]
or, in general
DSorxn = SnSo(products) - SnSo(reactants)
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To calculate DSrxn (which is DSsys), look up the
values on your reference sheets.
Calculate the standard entropy for the
formation of ammonia from nitrogen gas
and hydrogen gas at 25 oC.
N2(g) + 3H2(g)
2NH3(g)
DSorxn = SnSo(products) - SnSo(reactants)
DSorxn = (2 mol)(193 J/K.mol) - [(1 mol)(192 J/K.mol) + (3 mol)(131 J/K.mol)]
= -199 J/K
Does this reaction result in an increase or decrease in order?
Decrease because DS is negative
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Typically, if a reaction produces more
gas molecules than it consumes, DSo is
positive, likewise, if the total number
of gas molecules diminishes, DSo is
negative. If there is no change in the
total number of gas molecules, then
DSo may be positive or negative, but
will be relatively small numerically.
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Predict whether the entropy
change of the system in each
of the following reactions is
positive or negative.
(a) 2H2(g) + O2(g)
(b) NH4Cl(s)
2H2O(l)
NH3(g) + HCl(g)
(c) H2(g) + Br2(g)
2HBr(g)
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-DS
+DS
?DS
it will be
small
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Entropy Changes in the Surroundings
When an exothermic process takes place in the
system, the heat transferred to the
surroundings enhances motion of the
molecules in the surroundings. Consequently,
there is an increase in disorder of the
surroundings at the molecular level, and the
entropy of the surroundings increases.
Conversely, an endothermic process in the
system absorbs heat from the surroundings
and so decreases the entropy of the
surroundings because molecular motion
decreases.
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For constant-pressure processes the heat
change is equal to the enthalpy change of
the system. Therefore, the change in
entropy of the surroundings, DSsurr, is
proportional to DHsys.
DSsurr a -DHsys
The minus sign is used because if the
process is exothermic, DHsys is negative
and DSsurr is a positive quantity, indicating
an increase in entropy. On the other hand,
for an endothermic process, DHsys is
positive and the negative sign ensures that
the entropy of the surroundings decreases.
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The change in entropy for a given amount of
heat also depends on the Kelvin temperature.
If the temperature of the surroundings is
high, the molecules are already quite
energetic. Therefore, the absorption of heat
from an exothermic process in the system will
have relatively little impact on molecular
motion and the resulting increase in entropy
will be small. However, if the temperature of
the surroundings is low, then the addition of
the same amount of heat will cause a more
drastic increase in molecular motion and
hence a larger increase in entropy.
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From the inverse relationship between
DSsurr and temperature (in Kelvins) we
can rewrite the relationship between
DH, T and DS as
DSsurr = -DHsys
T
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Would you predict the synthesis of gaseous ammonia
from nitrogen gas and hydrogen gas to be
spontaneous at 25 oC? (Will it be +DSuniv ?)
DSuniv = DSsys + DSsurr
1/2 N2(g) + 3/2 H2(g)
NH3(g)
DH = -46.3 kJ or -46300 J
DSsys = (1)(193.0) - [(1/2)(191.5) + (3/2)(131.0)] = -99.3 J/K.mol
DSsurr = -DHsys
T
DSsurr = -(-46300 J) = 155 J/K
298 K
DSuniv = DSsys + DSsurr
DSuniv = -99.3 J/K.mol + 155 J/K.mol = 55.7 J/K
Because DSuniv is positive, we predict that
the reaction is spontaneous at 25 oC.
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It is important to keep in mind that
just because a reaction is
spontaneous does not mean that it
will occur at an observable rate.
Thermodynamics can tell us whether a
reaction will occur spontaneously
under specific conditions, but it does
not say how fast it will occur.
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Gibbs Free Energy
The second law of thermodynamics tells
us that a spontaneous reaction
increases the entropy in the universe
(+DS), but in order to determine the
sign of DSuniv, we would need to
calculate both DSsys and DSsurr. In
order to express the spontaneity of a
reaction more directly, we can use
another thermodynamic function
called Gibbs free energy (G).
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The change in free energy (DG) of a
system for a constant-temperature
process is
DG = DHsys - TDSsys
If a particular reaction is accompanied
by a release of usable energy (-DG),
the reaction will occur spontaneously,
if DG is equal to zero, the system is
at equilibrium.
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For the reaction
C(s) + O2(g)
CO2(g)
the values of DH and DS are known to
be -393.5 kJ and 3.05 J/K,
respectively. Would this reaction be
spontaneous at 25 oC?
DG = DH - TDS
= -393500 J - (298 K)(3.05 J/K)
= -394,000 J or -394 kJ
-DG, therefore the reaction would be
spontaneous
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The standard free-energy of reaction
(DGorxn) is the free-energy change for
a reaction when it occurs under
standard-state conditions, when
reactants in their standard states
are converted to products in their
standard states.
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To calculate DGorxn we start with the equation
aA + bB
cC + dD
the standard free-energy change for this
reaction is given by the equation
DGorxn = [cDGf(C) + dDGf(D)] - [aDGf(A) + bDGf(B)]
or, in general,
DGorxn = SnGf(products) - SnGf(reactants)
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Calculate the standard free-energy
changes for the combustion of 1 mol
of methane at 25 oC.
DGorxn = [1DGf(CO2) + 2DGf(H2O)] - [1DGf(CH4) + 2DGf(O2)]
DGorxn = [(1 mol)(-394.4 kJ/mol) + (2 mol)(-237.2 kJ/mol)]
- [(1 mol)(-50.8 kJ/mol) + (2 mol)(0 kJ/mol)]
DGorxn = -818.0 kJ
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Summarizing the conditions for spontaneity
and equilibrium at constant temperature
and pressure in terms of DG
DG < 0
DG > 0
DG = 0
The reaction is spontaneous in the
forward direction
The reaction is nonspontaneous.
The reaction is spontaneous in the
opposite direction.
The system is at equilibrium.
There is no net change.
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Temperature and Chemical
Reactions
The temperature at which a reaction
occurs (becomes spontaneous) is
important to the practical chemist.
We can make a reasonable estimate of
that temperature using the data on
your reference sheets.
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Calcium oxide (CaO) also called quicklime, is an
extremely valuable inorganic substance. It is
prepared by decomposing limestone (CaCO3) in a
kiln at a high temperature according to the
following reaction
CaCO3(s)
CaO(s) + CO2(g)
Estimate the temperature at which decomposition
becomes spontaneous.
DG = DH - TDS
DH = [(1 mol)(-635.6 kJ/mol) + (1 mol)(-393.5 kJ/mol)] - [(1 mol)(-1206.9 kJ/mol)]
DH = 177.8 kJ
DS = [(1 mol)(39.8 J/K.mol) + (1 mol)(213.6 J/K.mol)] - [(1 mol)(92.9 J/K.mol)]
DS = 130.0 J/K.mol
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DG = DH - TDS
DG = 177.8kJ - (298 K)(.1605 kJ/K)
DG = 130.0 kJ
Since DG is a large positive quantity, we conclude that the
reaction is not spontaneous at 25 oC. In order to make DG
negative, we first have to find the temperature at which
DG is zero, the point at which the system is at equilibrium.
At 835 oC the system is at
equilibrium. At temperatures
higher than 835 oC, DG
becomes negative, indicating
that the decomposition is
spontaneous. Reactions with
+DH and +DS are spontaneous
at high temps!
0 = DH - TDS
-DH = -TDS
DH = TDS
DH = T
DS
T = 177800 J
= 1108 K or 835 oC
160.5 J/K
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Two things need to be kept in mind regarding
the previous calculation. First, we used DH
and DS values at 25 oC to calculate changes
that occur at a much higher temperature, so
the value of DG will just be close to the
actual value. Second, some decomposition
will occur at below 835 oC (just like some
water will evaporate at temps below 100 oC.
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Phase Changes
Phase changes will occur when a system is at
equilibrium (DG = 0).
At approximately what temperature are the liquid
and gaseous bromine at equilibrium? (Or in
other words, estimate the normal boiling point
of liquid Br2.)
Br2(l)  Br2(g)
DH = 31.0 kJ/mol
DS = 93.0 J/K.mol
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DG = DH – TDS
0 = DH – TDS
DH = TDS
T = DH
DS
= 3.10 x 104 J/mol = 333 K
93.0 J/K . mol
So this means that at any temp ABOVE 333 K, the
above process will be spontaneous, therefore, 333 K is
Bromine’s boiling point.
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Free Energy and the
Equilibrium Constant
The temperature at which a reaction
occurs (becomes spontaneous) is
important to the practical chemist.
We can make a reasonable estimate of
that temperature using the data on
your reference sheets.
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We can use the value of DGo to
calculate the value of DG under
nonstandard conditions.
DG = DGo + RT lnQ
In this equation, R is the ideal gas
constant, 8.314 J/mol K, T is the
absolute temperature, and Q is the
reactant quotient.
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Under standard conditions, all the
reactants and products are equal to 1.
Thus, under standard conditions, Q = 1
and therefore, lnQ = 0, and DG = DGo,
as it should under standard
conditions.
When the concentrations of reactants
and products are nonstandard, we
must calculate the value of Q to
determine DG
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Calculate DG for the formation of
ammonia at 298 for a reaction
mixture that consists of 1.0 atm N2,
3.0 atm H2, and 0.50 atm NH3.
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Calculate DG at 298 for a reaction mixture
that consists of 1.0 atm N2, 3.0 atm H2,
and 0.50 atm NH3, DGo = -33.3 kJ
3H2 + N2  2NH3
(.50)2
(PNH3)2
-3
=
9.3
x
10
Q=
=
3
(1.0)(3.0)3
(PN2)(PH2)
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DG = DGo + RT lnQ
DG = -33.3 kJ/mol + (.008314 kJ/mol K)(298 K)ln(9.3 x 10-3)
DG = -33.3 kJ/mol + (-11.6 kJ/mol) = -44.9 kJ/mol
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We can use the previous equation to
derive the relationship between DGo
and the equilibrium constant, k, for
the reaction.
DG = DGo + RT lnQ
at equilibrium, DG = 0, and Q = k
0 = DGo + RT ln k
DGo = - RT ln k
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Given the value of DGo from the
previous calculation, solve for the
equilibrium constant for the
formation of ammonia at 298 K.
DGo = - RT ln k
-33,300 J/mol = - (8.314 J/mol K)(298 K) ln k
13.4 = ln k
Taking the inverse
ln of both sides…
6.60x 105 = k
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Which makes
sense, a –DG
should equal a
k > 1!
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The End
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